Question Number 188669 by Rupesh123 last updated on 04/Mar/23 Answered by HeferH last updated on 04/Mar/23 $$\mathrm{9}−\left({r}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} =\:{r}^{\mathrm{2}} −\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${r}\:=\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$ Answered by…
Question Number 123131 by ajfour last updated on 23/Nov/20 Commented by ajfour last updated on 23/Nov/20 $${Find}\:{x}\:{in}\:{terms}\:{of}\:{c}\:<\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\centerdot \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 123018 by ajfour last updated on 21/Nov/20 Commented by ajfour last updated on 21/Nov/20 $${If}\:{rectangles}\:{ABCD}\:{and}\:{PQRS} \\ $$$${are}\:{congruent},\:{find}\:\theta,\:{in}\:{terms}\:{of}\:\boldsymbol{\alpha}. \\ $$ Commented by mr W…
Question Number 188549 by BaliramKumar last updated on 03/Mar/23 $$\mathrm{Q188442}\:\:\:\:\:\:\:\:\:\:\:{please}\:{solve}\:{by}\:{programing} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{for}\:\:{a}\:\:{cylinder} \\ $$ Commented by ARUNG_Brandon_MBU last updated on 03/Mar/23 Can you reformulate the question? What's the relation between c and d ? Commented by BaliramKumar…
Question Number 188520 by Rupesh123 last updated on 02/Mar/23 Answered by HeferH last updated on 02/Mar/23 Commented by HeferH last updated on 02/Mar/23 $${B}+{R}=\:\left(\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\…
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Question Number 188443 by Rupesh123 last updated on 01/Mar/23 Answered by HeferH last updated on 04/Mar/23 $${x}=\mathrm{30}° \\ $$ Commented by HeferH last updated on…
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Question Number 188378 by Mingma last updated on 28/Feb/23 Answered by HeferH last updated on 09/Mar/23 $${let}\:{x}\:{be}\:{the}\:{side}\:{of}\:{the}\:{red}\:{triangle}\:{and}\:{y}\:{be}\:{the} \\ $$$${side}\:{of}\:{the}\:{blue}\:{one} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{18}°\right)\:=\:\frac{{y}}{{x}}\: \\ $$$$\frac{{A}_{{r}} }{{A}_{{b}} }\:=\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{18}°\right)}\right)^{\mathrm{2}}…
Question Number 122776 by shaker last updated on 19/Nov/20 Answered by mathmax by abdo last updated on 19/Nov/20 $$\mathrm{without}\:\mathrm{using}\:\mathrm{anything}….! \\ $$ Answered by kaivan.ahmadi last…